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Topic 11: Wave phenomena 11.3 Diffraction

Diffraction at a single slit 11.3.1 Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit. 11.3.2 Derive the formula  =   /b for the position of the first minimum of the diffraction pattern produced at a single slit.

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Topic 11: Wave phenomena 11.3 Diffraction

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  1. Diffraction at a single slit 11.3.1 Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit. 11.3.2 Derive the formula = /b for the position of the first minimum of the diffraction pattern produced at a single slit. 11.3.3 Solve problems involving single-slit diffraction. Be able to apply the formula  = /b. Topic 11: Wave phenomena11.3 Diffraction

  2. Diffraction at a single slit Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit. If a wave meets a hole in a wall that is of comparable size to its wavelength, the wave will be bent through a process called diffraction. If the aperture (hole, opening, etc.) is much larger than the wavelength, diffraction will be minimal to nonexistent. Topic 4: Oscillations and waves4.5 Wave properties REFLECTED WAVE DIFFRACTED WAVE INCIDENT WAVE

  3. Diffraction at a single slit Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit. Huygen’s principle states “Every point on a wavefront emits a spherical wavelet of the same velocity and wavelength as the original wave.” Note that because of Huygen’s principle waves can turn corners. Topic 4: Oscillations and waves4.5 Wave properties

  4. b b b Diffraction at a single slit Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit. The reason waves can turn corners is that the incoming wave transmits a disturbance by causing the medium to vibrate. And wherever the medium vibrates it becomes the center of a new wave front as illustrated. Note that the smaller the aperture b the more pronounced the effect. Topic 4: Oscillations and waves4.5 Wave properties b = 2 b = 6 b = 12

  5. Constructive interference Destructive interference Diffraction at a single slit Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit. Huygen’s wavelets not only allow the wave to turn corners, they also interfere with each other. Topic 4: Oscillations and waves4.5 Wave properties RELATIVE INTENSITY

  6. Diffraction at a single slit Sketch the variation with angle of diffraction of the relative intensity of light diffracted at a single slit. Topic 4: Oscillations and waves4.5 Wave properties EXAMPLE: If light is diffracted by a circular hole, a planar cross-section of the interference looks like the picture below. What will the light look like head-on?

  7. x b y Diffraction at a single slit Derive the formula = /b for the position of the first minimum of the diffraction pattern produced at a single slit. Consider a single slit of width b. From Huygen’s principle we know that every point within the slit acts as a wavelet. At the central maximum we see that the distance traveled by all the wavelets is about equal, and thus has constructive interference. Consider points x at one edge and y at the center of the slit: At the 1st minimum, the difference in distance (dashed) must be /2. Why? Topic 11: Wave phenomena11.3 Diffraction 1st min  Condition for destructive interference.

  8. 1st min  x b 2 x b  y y  2 Diffraction at a single slit Derive the formula = /b for the position of the first minimum of the diffraction pattern produced at a single slit. We will choose the midpoint of the slit (y) as our reference. And we will call the angle between the reference and the first minimum . We construct a right triangle as follows: Why does the side opposite  equal /2? Topic 11: Wave phenomena11.3 Diffraction   Condition for destructive interference.

  9. x  b 2  y  2  = /b ( in radians) location of first minimum in single slit diffraction Diffraction at a single slit Derive the formula = /b for the position of the first minimum of the diffraction pattern produced at a single slit. From the right triangle we see that sin = (/2)/(b/2) sin = /b. Perhaps you recall that if  is very small (and in radians) then sin   ( in rad). Finally… Topic 11: Wave phenomena11.3 Diffraction

  10. 10 0 -10 Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b. Topic 11: Wave phenomena11.3 Diffraction PRACTICE: Sketch in the diffraction patterns in the double-slit breakwater with 5-m waves. Then map out MAX (10 m), MIN (-10 m) and 0 m points. INCIDENT WAVE DIFFRACTED WAVE DIFFRACTED WAVE REFLECTED WAVE

  11. Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b. Topic 11: Wave phenomena11.3 Diffraction Diffraction allows waves to turn corners. All waves diffract-not just sound.

  12. Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b. Topic 11: Wave phenomena11.3 Diffraction Huygen says the wavelets will be spherical.

  13. Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b. Topic 11: Wave phenomena11.3 Diffraction

  14. Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b. Topic 11: Wave phenomena11.3 Diffraction Cent Caused by path length difference (PLD) along b. 2nd 1st 2nd 1st min  PLD = /2. 2nd min  PLD = 3/2. Central max  PLD = 0. 2nd max  PLD = .

  15. Diffraction at a single slit Solve problems involving single-slit diffraction. Be able to apply the formula  = /b. Topic 11: Wave phenomena11.3 Diffraction d  = /b tan  = d/D tan    for small . d =D/b.  = d/D =/b.

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